A bag contains 21 balls numbered 1 to 21. two balls are drawn one after another without replacement. Find the probability that both balls are even numbered.
(A) 3/14
(B) 2/3
(C) 6/9
(D) 1/8
Answer: (A)
Explanation:
Total numbers in sample space is 21
Probability of choosing any two of them is 21C2.
There are total 10 even numbers from 1 to 21 and probability of choosing any two of them is 10C2
Therefore the required answer is 10C2/21C2 = 3/14
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